Question

The amplitude of a simple harmonic oscillator is \(A\). When the velocity of particle is half of its maximum velocity, then its position is at

• A. \(\frac{A}{2}\)
• B. \(\frac{\sqrt{3}A}{4}\)
• C. \(\frac{A}{4}\)
• D. \(\frac{\sqrt{3}A}{2}\)

Hint:

In SHM, use \(v=\omega\sqrt{A^2-x^2}\). At mean position velocity is maximum, and at extreme position velocity is zero.

Answer 1

The Correct Option is D

For simple harmonic motion, velocity at displacement \(x\) is: \[ v=\omega\sqrt{A^2-x^2}. \] The maximum velocity is: \[ v_{\max}=\omega A. \] Given that the velocity is half of maximum velocity: \[ v=\frac{v_{\max}}{2}. \] So: \[ \omega\sqrt{A^2-x^2}=\frac{\omega A}{2}. \] Cancel \(\omega\) from both sides: \[ \sqrt{A^2-x^2}=\frac{A}{2}. \] Now square both sides: \[ A^2-x^2=\frac{A^2}{4}. \] Bring terms together: \[ x^2=A^2-\frac{A^2}{4}. \] \[ x^2=\frac{3A^2}{4}. \] Taking square root: \[ x=\frac{\sqrt{3}A}{2}. \] Hence, the position is: \[ \frac{\sqrt{3}A}{2}. \]